Question

What is the oxidation number of underlined element?

$H_2\underset{–}{S}{O}_4$

Answer 1

Let the oxidation number of S in $H_2\underset{–}{S}{O}_4$ be $x$.

Since the overall charge on the compound is $0$, the sum of oxidation states of all elements in it should be equal to $0$.

Therefore, $2(+1)+x+4(-2)=0$

or, $x=+6$

Hence, the oxidation number of $S$ in $H_2\underset{–}{S}{O}_4$ is $+6$.